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I'm new to the concept of complex plane. I found this exercise:

Let $z,z_1,z_2\in\mathbb C$ such that $z=z_1/z_2$. Show that the length of $z$ is the quotient of the length of $z_1$ and $z_2$.

If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $|z_1|=\sqrt{x_1^2+y_1^2}$ and $|z_2|=\sqrt{x_2^2+y_2^2}$, which yields $|z_1|/|z_2|=\sqrt{\dfrac{x_1^2+y_1^2}{x_2^2+y_2^2}}$.

Now, $z=\dfrac{z_1}{z_2}=\dfrac{x_1+iy_1}{x_2+iy_2} $. The first issue is to try and separate the imaginary part from the real one. I did this by: $$\dfrac{x_1+iy_1}{x_2+iy_2}=\dfrac{x_1+iy_1}{x_2+iy_2}\times\frac{x_2-iy_2}{x_2-iy_2}=\frac{x_1x_2-ix_1y_2+ix_2y_1+y_1y_2}{x_2^2+y_2^2}\\ =\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+i\frac{x_2y_1-x_1y_2}{x_2^2+y_2^2}.$$ Hence $|z|=\sqrt{\left(\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\right)^2+\left(\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}\right)^2}=\sqrt{\dfrac{(x_1x_2+y_1y_2)^2+(x_2y_1-x_1y_2)^2}{(x_2^2+y_2^2)^2}}$. Continuing I get $$\sqrt{\frac{(x_1x_2)^2+(y_1y_2)^2+(x_2y_1)^2+(x_1y_2)^2}{(x_2^2+y_2^2)^2}},$$ but I can't see any future here. Is there any mistake? How can I simplify the expression for $|z|$? I appreciate your help.

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  • $\begingroup$ The top line of the bottom line is $(x_1^2+y_1^2)(x_2^2+y_2^2)$. $\endgroup$ – JMP Feb 7 '15 at 1:34
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Given $z = z_1/z_2$, you can conclude that $z z_2 = z_1$, so $|z_1| = |z z_2|$. If you can show that $|z z_2| = |z||z_2|$ then you can divide both sides by $|z_2|$ to get the desired result.

So you just need to know that the modulus of a product of two numbers is the product of the modulus of each number.

You can write any complex number $z$ as $z = r e^{i\theta}$ where $r = |z|$ and $\theta = \mathrm{Arg}\ z$. This works in reverse, too: if $r$ and $\theta$ are real and $r e^{i\theta} = z$, then $|z| = r$.

So for a product of two numbers $z_a z_b$, let $z_a = r_a e^{i\theta_a}$ and $z_b = r_b e^{i\theta_b}.$ Then $$z_a z_b = r_a e^{i\theta_a} r_b e^{i\theta_b} = r_a r_b e^{i(\theta_a+\theta_b)},$$ and so $|z_a z_b| = r_a r_b = |z_a||z_b|.$

You could also have proved the ratio formula directly in this fashion, but I find multiplication easier to work with.

In terms of developing the theory, maybe it's too early to start using the exponential format,
but it's a handy way to remember facts like this.

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Since $|z|^2=z\overline{z}$,

$$\left|{z_1\over z_2}\right|^2={z_1\over z_2}{\overline{z_1}\over\overline{z_2}}={|z_1|^2\over|z_2|^2}.$$

Take square root of both sides to get your result.

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  • $\begingroup$ Thanks, I'll take a look at complex conjugate :P $\endgroup$ – Vladimir Vargas Feb 6 '15 at 22:43
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If you know that $|ab|=|a|\,|b|$, then use $a=z_1/z_2$ and $b=z_2$: $$ |z_1|=\left|\frac{z_1}{z_2}z_2\right|=|ab|=|a|\,|b|= \left|\frac{z_1}{z_2}\right|\,|z_2| $$ Now, easily, $$ \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}. $$

Proving that $|ab|=|a|\,|b|$ is much simpler, even with the definition: if $a=x_1+iy_1$ and $b=x_2+iy_2$, then $$ ab=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1) $$ and it's quite easy to verify that $$ (x_1x_2-y_1y_2)^2+(x_1y_2+x_2y_1)^2=(x_1^2+y_1^2)(x_2^2+y_2^2) $$

However, it's much easier with the definition $$ |z|=\sqrt{z\bar{z}}. $$

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  • $\begingroup$ Yes, thanks for your help $\endgroup$ – Vladimir Vargas Feb 6 '15 at 23:00
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Let $z_1=r.e^{i\theta}$ and $z_2=R.e^{i\alpha}$. The result follows easily.

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